Question

Consider a system of two masses and a pulley shown in the figure. The coefficient of friction between the two blocks and also between block and table is $\mathrm{0.1.}$ Find the force $F$, that must be given to the $0.8kg$ block such that it attains acceleration of $5m/s^2$. (Assume, acceleration due to gravity, $g=10m/s^2.)$

• A. 6.4 N
• B. 7.1 N
• C. 6.0 N
• D. 7.8 N

Answer 1

Answer: (A)

Given, acceleration of block, $a=5m/s^2$ and coefficient of friction between two blocks and table, $\mu =0.1$

Free body diagram for $0.8kg$ block.

For $0.8kg$ block, friction force due to $0.2kg$ block is in opposite direction of applied force $F$.
$\therefore$ Equation of motion for $0.8kg$ block,
$F-T-\mu R-\mu R_1=0.8a$
putting the given values, we get
$F-T-0.1(0.8+0.2)g-0.1\times 0.2g=0.8a$
$F-T-0.1\times 10-0.1\times 0.2\times 10=0.8\times 5$
$F-T=5.2...(i)$
Free body diagram for $0.2kg$ body,

For $0.2kg$ block, friction force due to $0.8kg$ block is in opposite direction of tension force.
So, $T-\mu R_1=0.2a$
$T-0.1\times 0.2g=0.2\times 5$
$T-0.02\times 10=1.0$
$T=1.2...(ii)$
From Eqs. (i) and (ii), we get
$F-1.2=5.2$
$F=5.2+1.2=6.4N$